Q03 Test 26 English

Test No. 26 – Complete Solutions

103

Convert 73.12° into degrees, minutes, and seconds.

🔑 Key Concept: The decimal part of a degree is converted to minutes by multiplying by 60, and the decimal part of the resulting minutes is converted to seconds by multiplying by 60 again.
\(1° = 60' \quad\) and \(\quad 1' = 60''\)

Given: \(73.12°\)

Step 1 – Separate the whole degrees: Whole degrees \(= 73°\)
Decimal part \(= 0.12°\)

Step 2 – Convert decimal degrees to minutes: \[0.12° \times 60 = 7.2'\] Whole minutes \(= 7'\), decimal part \(= 0.2'\)

Step 3 – Convert decimal minutes to seconds: \[0.2' \times 60 = 12''\]

\(\boxed{73.12° = 73°\ 7'\ 12''}\)

104

Convert radians to degrees: (i) 1.2 rad    (ii) \(\dfrac{11\pi}{6}\) rad

🔑 Key Concept: Use the fundamental radian–degree relationship: \[1 \text{ radian} = \frac{180°}{\pi} \approx 57.2958°\] So: \(\theta_{\text{deg}} = \theta_{\text{rad}} \times \dfrac{180°}{\pi}\)

(i) Convert 1.2 rad to degrees

\[1.2 \times \frac{180°}{\pi} = \frac{216°}{\pi}\]

Using \(\pi \approx 3.14159\): \[\frac{216}{3.14159} \approx 68.7549°\]

Convert decimal to minutes & seconds: \(0.7549° \times 60 = 45.294'\)
\(0.294' \times 60 \approx 17.6'' \approx 18''\)

\(\boxed{1.2 \text{ rad} \approx 68°\ 45'\ 18'' \approx 68.75°}\)

(ii) Convert \(\dfrac{11\pi}{6}\) rad to degrees

\[\frac{11\pi}{6} \times \frac{180°}{\pi} = \frac{11 \times 180°}{6} = \frac{1980°}{6} = 330°\] (The \(\pi\) cancels — this gives an exact result!)

\(\boxed{\dfrac{11\pi}{6} \text{ rad} = 330°}\)

105

Find the distance between \(B(5,\,-2)\) and \(A(5,\,6)\).

🔑 Key Concept – Distance Formula: \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\] When the x-coordinates are equal, the points lie on a vertical line, and the distance simplifies to the absolute difference of the y-coordinates: \[d = |y_2 - y_1|\]

Identify coordinates: \(B(x_1, y_1) = (5,\,-2)\) and \(A(x_2, y_2) = (5,\,6)\)

Apply the distance formula: \[d = \sqrt{(5-5)^2 + (6-(-2))^2}\] \[= \sqrt{(0)^2 + (8)^2}\] \[= \sqrt{0 + 64} = \sqrt{64}\]

Note: Since \(x_1 = x_2 = 5\), both points lie on the vertical line \(x = 5\). The distance equals \(|6-(-2)| = |8| = 8\).

\(\boxed{d = 8 \text{ units}}\)

106

Find the unknown quantities in the following figures (similar triangles).

🔑 Key Concept – Ratio of Areas of Similar Triangles: For two similar triangles with corresponding sides \(l_1\) and \(l_2\) and areas \(A_1\) and \(A_2\): \[\frac{A_1}{A_2} = \left(\frac{l_1}{l_2}\right)^2\] This formula states that the ratio of areas equals the square of the ratio of corresponding sides. Rearranging allows us to find any unknown area or side length.

Figure 1 — Find Area \(A_1\)

Given: Triangle 1 has corresponding side \(l_1 = 2.4\text{ cm}\).
Triangle 2 has corresponding side \(l_2 = 1.5\text{ cm}\) and area \(A_2 = 25\text{ cm}^2\).

Step 1 – Write the formula: \[\frac{A_1}{A_2} = \left(\frac{l_1}{l_2}\right)^2\]

Step 2 – Substitute known values: \[\frac{A_1}{25} = \left(\frac{2.4}{1.5}\right)^2\]

Step 3 – Simplify the ratio and square it: \[\frac{2.4}{1.5} = 1.6\] \[\frac{A_1}{25} = (1.6)^2 = 2.56\]

Step 4 – Solve for \(A_1\): \[A_1 = 2.56 \times 25 = 64 \text{ cm}^2\]

\(\boxed{A_1 = 64 \text{ cm}^2}\)